Three Canadians, 4 Americans, and 2 Mexicans attend a trade conference. In how many ways can they be seated in a row if the people of the same nationality are to be seated next to each other?
We calculate the number of ways each nationality can order their members. This is simple permutations.
Canadians: 3! = 6
Americans: 4! = 12
Mexicans: 2! = 2
However, we can also manipulate the order of our diplomats (i.e. Canada, America, Mexico v. Canada, Mexico, America).
Again since, there are 3 options, we can simply calculate a permutation, 3! = 6.
We now take the sum of the first part and multiply it by the second part.
(6 + 12 + 2) * 6 = 120
To solve this problem, we can consider the people from each nationality as distinct entities. Let's start by calculating the number of ways we can arrange the nationalities.
1. Arrange the Canadians: Since there are 3 Canadians, they can be seated amongst themselves in 3! = 6 ways.
2. Arrange the Americans: Since there are 4 Americans, they can be seated amongst themselves in 4! = 24 ways.
3. Arrange the Mexicans: Since there are 2 Mexicans, they can be seated amongst themselves in 2! = 2 ways.
Now, to calculate the total number of arrangements, we multiply the number of ways each group can be seated:
Total arrangements = 6 * 24 * 2 = 288
So, there are 288 ways to seat the group if people of the same nationality are seated next to each other.